Trigonometry Worksheets - Printable Worksheet for Trigonometry ... - Free Printable

To solve the problems in the image, we need to use trigonometric relationships (sine, cosine, and tangent) to find the missing lengths. Let's go through each problem step by step.

---

Section A

#### 1)
Given:
- Hypotenuse = 12 cm
- Angle = 30°
- Side opposite the angle \( x \)

Using the sine function:
\[
\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
\[
\sin(30^\circ) = \frac{x}{12}
\]
\[
\sin(30^\circ) = 0.5
\]
\[
0.5 = \frac{x}{12}
\]
\[
x = 12 \times 0.5 = 6 \text{ cm}
\]

#### 2)
Given:
- Adjacent side = 9 cm
- Angle = 67°
- Hypotenuse \( x \)

Using the cosine function:
\[
\cos(67^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}
\]
\[
\cos(67^\circ) = \frac{9}{x}
\]
\[
\cos(67^\circ) \approx 0.391
\]
\[
0.391 = \frac{9}{x}
\]
\[
x = \frac{9}{0.391} \approx 23.0 \text{ cm}
\]

#### 3)
Given:
- Opposite side = 16 cm
- Angle = 40°
- Adjacent side \( x \)

Using the tangent function:
\[
\tan(40^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(40^\circ) = \frac{16}{x}
\]
\[
\tan(40^\circ) \approx 0.839
\]
\[
0.839 = \frac{16}{x}
\]
\[
x = \frac{16}{0.839} \approx 19.1 \text{ cm}
\]

#### 4)
Given:
- Hypotenuse = 17 cm
- Angle = 52°
- Adjacent side \( x \)

Using the cosine function:
\[
\cos(52^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}
\]
\[
\cos(52^\circ) = \frac{x}{17}
\]
\[
\cos(52^\circ) \approx 0.616
\]
\[
0.616 = \frac{x}{17}
\]
\[
x = 17 \times 0.616 \approx 10.5 \text{ cm}
\]

#### 5)
Given:
- Adjacent side = 6 cm
- Angle = 71°
- Opposite side \( x \)

Using the tangent function:
\[
\tan(71^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(71^\circ) = \frac{x}{6}
\]
\[
\tan(71^\circ) \approx 2.904
\]
\[
2.904 = \frac{x}{6}
\]
\[
x = 6 \times 2.904 \approx 17.4 \text{ cm}
\]

#### 6)
Given:
- Opposite side = 21 cm
- Angle = 37°
- Hypotenuse \( x \)

Using the sine function:
\[
\sin(37^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
\[
\sin(37^\circ) = \frac{21}{x}
\]
\[
\sin(37^\circ) \approx 0.602
\]
\[
0.602 = \frac{21}{x}
\]
\[
x = \frac{21}{0.602} \approx 34.9 \text{ cm}
\]

---

Section B

#### 1)
Given:
- Adjacent side = 14 cm
- Angle = 15°
- Opposite side \( x \)

Using the tangent function:
\[
\tan(15^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(15^\circ) = \frac{x}{14}
\]
\[
\tan(15^\circ) \approx 0.268
\]
\[
0.268 = \frac{x}{14}
\]
\[
x = 14 \times 0.268 \approx 3.75 \text{ cm}
\]

#### 2)
Given:
- Opposite side = 19 cm
- Angle = 71°
- Adjacent side \( x \)

Using the tangent function:
\[
\tan(71^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(71^\circ) = \frac{19}{x}
\]
\[
\tan(71^\circ) \approx 2.904
\]
\[
2.904 = \frac{19}{x}
\]
\[
x = \frac{19}{2.904} \approx 6.54 \text{ cm}
\]

#### 3)
Given:
- Adjacent side = 5 cm
- Angle = 42°
- Opposite side \( x \)

Using the tangent function:
\[
\tan(42^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(42^\circ) = \frac{x}{5}
\]
\[
\tan(42^\circ) \approx 0.900
\]
\[
0.900 = \frac{x}{5}
\]
\[
x = 5 \times 0.900 = 4.50 \text{ cm}
\]

#### 4)
Given:
- Opposite side = 11 cm
- Angle = 38°
- Hypotenuse \( x \)

Using the sine function:
\[
\sin(38^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
\[
\sin(38^\circ) = \frac{11}{x}
\]
\[
\sin(38^\circ) \approx 0.616
\]
\[
0.616 = \frac{11}{x}
\]
\[
x = \frac{11}{0.616} \approx 17.9 \text{ cm}
\]

#### 5)
Given:
- Adjacent side = 7 cm
- Angle = 20°
- Opposite side \( x \)

Using the tangent function:
\[
\tan(20^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(20^\circ) = \frac{x}{7}
\]
\[
\tan(20^\circ) \approx 0.364
\]
\[
0.364 = \frac{x}{7}
\]
\[
x = 7 \times 0.364 \approx 2.55 \text{ cm}
\]

#### 6)
Given:
- Adjacent side = 24 cm
- Angle = 34°
- Opposite side \( x \)

Using the tangent function:
\[
\tan(34^\circ) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(34^\circ) = \frac{x}{24}
\]
\[
\tan(34^\circ) \approx 0.675
\]
\[
0.675 = \frac{x}{24}
\]
\[
x = 24 \times 0.675 \approx 16.2 \text{ cm}
\]

---

Final Answers

\[
\boxed{
\begin{array}{ll}
\text{Section A:} & \\
1. & x = 6.00 \text{ cm} \\
2. & x = 23.0 \text{ cm} \\
3. & x = 19.1 \text{ cm} \\
4. & x = 10.5 \text{ cm} \\
5. & x = 17.4 \text{ cm} \\
6. & x = 34.9 \text{ cm} \\
\text{Section B:} & \\
1. & x = 3.75 \text{ cm} \\
2. & x = 6.54 \text{ cm} \\
3. & x = 4.50 \text{ cm} \\
4. & x = 17.9 \text{ cm} \\
5. & x = 2.55 \text{ cm} \\
6. & x = 16.2 \text{ cm} \\
\end{array}
}
\]